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372 ro, 0, to 0.6357 0.3273 362.4 365 Continued \u2026.. PROBLEM 5.92 (Cont.) The temperatures from the one-term series calculations of the Example 5.7 are systematically higher than those resulting from the IHT multiple-term series model, which is the more accurate method. (b) The temperature histories for the center and mid-height of the side locations are shown in the graph below. Note that at early times, the temperature difference between these locations, and hence the gradient, is large. Large differences could cause variations in microstructure and hence, mechanical properties, as well as induce residual thermal stresses. (c) Effect of doubling the convection coefficient is to increase the quenching rate, but much less than by a factor of two as can be seen in the graph below. Quenching with h = 500 W/m^2.K 0 2 4 6 8 10 Time, t (min) 300 400 500 600 T(x , r, t) (C ) Mid-height of side (0,ro) Center (0, 0) Effect of increased conv. coeff. on quenching rate 0 2 4 6 8 10 Time, t (min) 300 400 500 600 T( 0, 0, t) ( C) h = 500 W/m^2.K h = 1000 W/m^2.K COMMENTS: From IHT menu for Transient Conduction | Plane Wall and Cylinder, the models were combined to solve the product solution. Key portions of the code, less the input variables, are copied below. // Plane wall temperature distribution // The temperature distribution is T_xtP = T_xt_trans("Plane Wall",xstar,FoP,BiP,Ti,Tinf) // Eq 5.39 // The dimensionless parameters are xstar = x / L BiP = h * L / k // Eq 5.9 FoP= alpha * t / L^2 // Eq 5.33 alpha = k/ (rho * cp) // Dimensionless representation, P(x,t) P_xt = (T_xtP - Tinf ) / (Ti - Tinf) // Cylinder temperature distribution // The temperature distribution T(r,t) is T_rtC = T_xt_trans("Cylinder",rstar,FoC,BiC,Ti,Tinf) // Eq 5.47 // The dimensionless parameters are rstar = r / ro BiC = h * ro / k FoC= alpha * t / ro^2 // Dimensionless representation, C(r,t) C_rt= (T_rtC - Tinf ) / (Ti - Tinf) // Product solution temperature distribution (T_xrt - Tinf) / (Ti - Tinf) = P_xt * C_rt PROBLEM 5.93 KNOWN: Stability criterion for the explicit method requires that the coefficient of the pmT term of the one-dimensional, finite-difference equation be zero or positive. FIND: For Fo > 1/2, the finite-difference equation will predict values of p+1mT which violate the Second law of thermodynamics. Consider the prescribed numerical values. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction in x, (2) Constant properties, (3) No internal heat generation. ANALYSIS: The explicit form of the finite-difference equation, Eq. 5.73, for an interior node is ( ) ( )p+1 p p pm mm+1 m-1T Fo T T 1 2 Fo T .= + + \u2212 The stability criterion requires that the coefficient of pmT be zero or greater. That is, ( ) 11 2 Fo 0 or Fo . 2 \u2212 \u2265 \u2264 For the prescribed temperatures, consider situations for which Fo = 1, ½ and ¼ and calculate p+1 mT . ( ) ( ) ( ) ( ) ( ) ( ) p+1 m p+1 m p+1 m Fo 1 T 1 100 100 C 1 2 1 50 C 250 C Fo 1/2 T 1 2 100 100 C 1 2 1/ 2 50 C 100 C Fo 1/4 T 1/ 4 100 100 C 1 2 1/ 4 50 C 75 C. = = + + \u2212 × = = = + + \u2212 × = = = + + \u2212 × = / $ $ $ $ $ $ $ $ $ Plotting these distributions above, note that when Fo = 1, p+1mT is greater than 100°C, while for Fo = ½ and ¼ , p+1mT \u2264 100°C. The distribution for Fo = 1 is thermodynamically impossible: heat is flowing into the node during the time period \u2206t, causing its temperature to rise; yet heat is flowing in the direction of increasing temperature. This is a violation of the Second law. When Fo = ½ or ¼, the node temperature increases during \u2206t, but the temperature gradients for heat flow are proper. This will be the case when Fo \u2264 ½, verifying the stability criterion. PROBLEM 5.94 KNOWN: Thin rod of diameter D, initially in equilibrium with its surroundings, Tsur, suddenly passes a current I; rod is in vacuum enclosure and has prescribed electrical resistivity, \u3c1e, and other thermophysical properties. FIND: Transient, finite-difference equation for node m. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional, transient conduction in rod, (2) Surroundings are much larger than rod, (3) Properties are constant and evaluated at an average temperature, (4) No convection within vacuum enclosure. ANALYSIS: The finite-difference equation is derived from the energy conservation requirement on the control volume, Ac\u2206x, where 2cA D / 4 and P D.\u3c0 \u3c0= = The energy balance has the form p+1 p 2 m m in out g st a b rad e T TE E E E q q q I R cV . t \u3c1 \u2212\u2212 + = + \u2212 + = \u2206 \ufffd \ufffd \ufffd \ufffd where 2g eE I R=\ufffd and e e cR x/A .\u3c1= \u2206 Using Fourier\u2019s law to express the conduction terms, qa and qb, and Eq. 1.7 for the radiation exchange term, qrad, find ( )p p p p p+1 pm m 4,p 4 2 e m mm-1 m+1c c m sur c c T T T T x T TkA kA P x T T I cA x . x x A t \u3c1 \u3b5 \u3c3 \u3c1 \u2212 \u2212 \u2206 \u2212 + \u2212 \u2206 \u2212 + = \u2206 \u2206 \u2206 \u2206 Divide each term by \u3c1cAc \u2206x/\u2206t, solve for p+1mT and regroup to obtain ( )p+1 p p pm mm-1 m+12 2k t k tT T T 2 1 T c cx x\u3c1 \u3c1 \uf8ee \uf8f9\u2206 \u2206 = \u22c5 + \u2212 \u22c5 \u22c5 \u2212\uf8ef \uf8fa \u2206 \u2206\uf8f0 \uf8fb ( ) 24,p 4 em sur 2 c c I P t tT T . A c cA \u3c1\u3b5 \u3c3 \u3c1 \u3c1 \u2206 \u2206 \u2212 \u22c5 \u2212 + \u22c5 Recognizing that Fo = \u3b1 \u2206t/\u2206x2, regroup to obtain ( ) ( ) ( ) 2 22p+1 p p p 4,p 4 em m m surm-1 m+1 2 c c I xP xT Fo T T 1 2 Fo T Fo T T Fo. kA kA \u3c1\u3b5 \u3c3 \u2206\u2206 = + + \u2212 \u2212 \u22c5 \u2212 + \u22c5 The stability criterion is based upon the coefficient of the pmT term written as Fo \u2264 ½. < COMMENTS: Note that we have used the forward-difference representation for the time derivative; see Section 5.9.1. This permits convenient treatment of the non-linear radiation exchange term. PROBLEM 5.95 KNOWN: One-dimensional wall suddenly subjected to uniform volumetric heating and convective surface conditions. FIND: Finite-difference equation for node at the surface, x = -L. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional transient conduction, (2) Constant properties, (3) Uniform \ufffdq. ANALYSIS: There are two types of finite-difference equations for the explicit and implicit methods of solution. Using the energy balance approach, both types will be derived. Explicit Method. Perform an energy balance on the surface node shown above, p+1 p o o in out g st conv cond T TE E E E q q qV cV t \u3c1 \u2212\u2212 + = + + = \u2206 \ufffd \ufffd \ufffd \ufffd \ufffd (1) ( )( ) ( ) p p p+1 pop o o1o T T T Tx xh 1 1 T T k 1 1 q 1 1 c 1 1 . x 2 2 t \u3c1 \u221e \u2212 \u2212\u2206 \u2206\uf8ee \uf8f9 \uf8ee \uf8f9 \u22c5 \u2212 + \u22c5 + \u22c5 \u22c5 = \u22c5 \u22c5\uf8ef \uf8fa \uf8ef \uf8fa\u2206 \u2206\uf8f0 \uf8fb \uf8f0 \uf8fb\ufffd (2) For the explicit method, the temperatures on the LHS are evaluated at the previous time (p). The RHS provides a forward-difference approximation to the time derivative. Divide Eq. (2) by \u3c1c\u2206x/2\u2206t and solve for p+1oT . ( ) ( )p+1 p p p po o o o12h t k t tT 2 T T 2 T T q T . c x c c x\u3c1 \u3c1\u3c1 \u221e \u2206 \u2206 \u2206 = \u2212 + \u2212 + + \u2206 \u2206 \ufffd Introducing the Fourier and Biot numbers, ( ) 2Fo k/ c t/ x Bi h x/k\u3c1\u2261 \u2206 \u2206 \u2261 \u2206 ( ) 2p+1 p p o o1 q xT 2 Fo T Bi T 1 2 Fo 2 Fo Bi T . 2k\u221e \uf8ee \uf8f9\u2206 = + \u22c5 + + \u2212 \u2212 \u22c5\uf8ef \uf8fa\uf8ef \uf8fa\uf8f0 \uf8fb \ufffd (3) The stability criterion requires that the coefficient of poT be positive. That is, ( ) ( )1 2 Fo 2 Fo Bi 0 or Fo 1/2 1 Bi .\u2212 \u2212 \u22c5 \u2265 \u2264 + (4) < Implicit Method. Begin as above with an energy balance. In Eq. (2), however, the temperatures on the LHS are evaluated at the new (p+1) time. The RHS provides a backward- difference approximation to the time derivative. ( ) p+1 p+1 p+1 pop+1 o o1o T T T Tx xh T T k q c x 2 2 t \u3c1 \u221e \u2212 \u2212\u2206 \u2206\uf8ee \uf8f9 \uf8ee \uf8f9 \u2212 + + =\uf8ef \uf8fa \uf8ef \uf8fa\u2206 \u2206\uf8f0 \uf8fb \uf8f0 \uf8fb\ufffd (5) ( )( ) 2p+1 p+1 po o1 q x1 2 Fo Bi 1 T 2 Fo T T 2Bi Fo T Fo .k\u221e \u2206 + + \u2212 \u22c5 = + \u22c5 \u22c5 + \ufffd (6) < COMMENTS: Compare these results (Eqs. 3, 4 and 6) with the appropriate expression in Table 5.2. PROBLEM 5.96 KNOWN: Plane wall,